行列の計算例題（行列式（余因子展開））

　　　　　　　　　　


例題：次の行列式を余因子展開を使って計算せよ。

\[ \left|\,\begin{array}{@{}rwr{20pt}wr{20pt}wr{20pt}@{}}{-}1 & 0 & 1 & {-}1 \\ 1 & 1 & 1 & {-}1 \\ 0 & 2 & 2 & 1 \\ {-}2 & {-}1 & 0 & 0\end{array}\,\right| \]

解答

\( \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}wr{15pt}@{}}{-}1 & 0 & 1 & {-}1 \\ 1 & 1 & 1 & {-}1 \\ 0 & 2 & 2 & 1 \\ {-}2 & {-}1 & 0 & 0\end{array}\,\right| \qquad \) （余因子展開） 第4列で展開する : [-1, -1, 1, 0]

\( \qquad\quad = (-1)^{4-1}\times(-1)\times\left|\,\begin{array}{@{}rwr{15pt}wr{15pt}@{}}1 & 1 & 1 \\ 0 & 2 & 2 \\ {-}2 & {-}1 & 0\end{array}\,\right| + (-1)^{4-2}\times(-1)\times\left|\,\begin{array}{@{}rwr{15pt}wr{15pt}@{}}{-}1 & 0 & 1 \\ 0 & 2 & 2 \\ {-}2 & {-}1 & 0\end{array}\,\right| + (-1)^{4-3}\times1\times\left|\,\begin{array}{@{}rwr{15pt}wr{15pt}@{}}{-}1 & 0 & 1 \\ 1 & 1 & 1 \\ {-}2 & {-}1 & 0\end{array}\,\right| \)

\( \qquad\quad = (-1)^{4-1}\times(-1)\times2 + (-1)^{4-2}\times(-1)\times2 + (-1)^{4-3}\times1\times0 \)

\( \qquad\quad = 0 \)

\( \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}@{}}1 & 1 & 1 \\ 0 & 2 & 2 \\ {-}2 & {-}1 & 0\end{array}\,\right| \qquad \) （余因子展開） 第3列で展開する : [1, 2, 0]

\( \qquad\quad = (-1)^{3-1}\times1\times\left|\,\begin{array}{@{}rwr{15pt}@{}}0 & 2 \\ {-}2 & {-}1\end{array}\,\right| + (-1)^{3-2}\times2\times\left|\,\begin{array}{@{}rwr{15pt}@{}}1 & 1 \\ {-}2 & {-}1\end{array}\,\right| \)

\( \qquad\quad = (-1)^{3-1}\times1\times4 + (-1)^{3-2}\times2\times1 \)

\( \qquad\quad = 2 \)

\( \qquad \left|\,\begin{array}{@{}rwr{15pt}@{}}0 & 2 \\ {-}2 & {-}1\end{array}\,\right| = 0\times(-1) - 2\times(-2) = 4 \)

\( \qquad \left|\,\begin{array}{@{}rwr{15pt}@{}}1 & 1 \\ {-}2 & {-}1\end{array}\,\right| = 1\times(-1) - 1\times(-2) = 1 \)

\( \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}@{}}{-}1 & 0 & 1 \\ 0 & 2 & 2 \\ {-}2 & {-}1 & 0\end{array}\,\right| \qquad \) （余因子展開） 第3列で展開する : [1, 2, 0]

\( \qquad\quad = (-1)^{3-1}\times1\times\left|\,\begin{array}{@{}rwr{15pt}@{}}0 & 2 \\ {-}2 & {-}1\end{array}\,\right| + (-1)^{3-2}\times2\times\left|\,\begin{array}{@{}rwr{15pt}@{}}{-}1 & 0 \\ {-}2 & {-}1\end{array}\,\right| \)

\( \qquad\quad = (-1)^{3-1}\times1\times4 + (-1)^{3-2}\times2\times1 \)

\( \qquad\quad = 2 \)

\( \qquad \left|\,\begin{array}{@{}rwr{15pt}@{}}0 & 2 \\ {-}2 & {-}1\end{array}\,\right| = 0\times(-1) - 2\times(-2) = 4 \)

\( \qquad \left|\,\begin{array}{@{}rwr{15pt}@{}}{-}1 & 0 \\ {-}2 & {-}1\end{array}\,\right| = (-1)\times(-1) - 0\times(-2) = 1 \)

\( \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}@{}}{-}1 & 0 & 1 \\ 1 & 1 & 1 \\ {-}2 & {-}1 & 0\end{array}\,\right| \qquad \) （余因子展開） 第3列で展開する : [1, 1, 0]

\( \qquad\quad = (-1)^{3-1}\times1\times\left|\,\begin{array}{@{}rwr{15pt}@{}}1 & 1 \\ {-}2 & {-}1\end{array}\,\right| + (-1)^{3-2}\times1\times\left|\,\begin{array}{@{}rwr{15pt}@{}}{-}1 & 0 \\ {-}2 & {-}1\end{array}\,\right| \)

\( \qquad\quad = (-1)^{3-1}\times1\times1 + (-1)^{3-2}\times1\times1 \)

\( \qquad\quad = 0 \)

\( \qquad \left|\,\begin{array}{@{}rwr{15pt}@{}}1 & 1 \\ {-}2 & {-}1\end{array}\,\right| = 1\times(-1) - 1\times(-2) = 1 \)

\( \qquad \left|\,\begin{array}{@{}rwr{15pt}@{}}{-}1 & 0 \\ {-}2 & {-}1\end{array}\,\right| = (-1)\times(-1) - 0\times(-2) = 1 \)